51NOD1522&CF567F 上下序列/Mausoleum DP

  1 #include<bits/stdc++.h>
  2 #define int long long
  3 using namespace std;
  4 
  5 int ans[72][72] , numOp[72];
  6 
  7 vector < int > num[72] , op[72];
  8 map < string , int > opToNum;
  9 
 10 bool ifOK(int l , int r , int que1 , int que2){
 11 //特别注意判断que1位置与que2位置有联系的情况
 12     for(int i = 0 ; i < numOp[que1] ; i++)
 13         switch(op[que1][i]){
 14             case 1:
 15                 if(num[que1][i] < r && num[que1][i] > l || num[que1][i] == que2)
 16                     return 0;
 17                 break;
 18             case 2:
 19                 if(num[que1][i] < r && num[que1][i] > l && num[que1][i] != que2)
 20                     return 0;
 21                 break;
 22             case 3:
 23                 if(!(num[que1][i] == que2))
 24                     return 0;
 25                 break;
 26             case 4:
 27                 if(!(num[que1][i] <= r && num[que1][i] >= l || num[que1][i] == que2))
 28                     return 0;
 29                 break;
 30             case 5:
 31                 if(!(num[que1][i] < r && num[que1][i] > l) || num[que1][i] == que2)
 32                     return 0;
 33         }
 34     for(int i = 0 ; i < numOp[que2] ; i++)
 35         switch(op[que2][i]){
 36             case 1:
 37                 if(num[que2][i] < r && num[que2][i] > l || num[que2][i] == que1)
 38                     return 0;
 39                 break;
 40             case 2:
 41                 if(num[que2][i] < r && num[que2][i] > l && num[que2][i] != que1)
 42                     return 0;
 43                 break;
 44             case 3:
 45                 if(!(num[que2][i] == que1))
 46                     return 0;
 47                 break;
 48             case 4:
 49                 if(!(num[que2][i] <= r && num[que2][i] >= l || num[que2][i] == que1))
 50                     return 0;
 51                 break;
 52             case 5:
 53                 if(!(num[que2][i] < r && num[que2][i] > l) || num[que2][i] == que1)
 54                     return 0;
 55         }
 56     return 1;
 57 }
 58 
 59 main(){
 60     opToNum.insert(make_pair(">" , 1));
 61     opToNum.insert(make_pair(">=" , 2));
 62     opToNum.insert(make_pair("=" , 3));
 63     opToNum.insert(make_pair("<=" , 4));
 64     opToNum.insert(make_pair("<" , 5));
 65     int N , K;
 66     for(cin >> N >> K ; K ; K--){
 67         int a , b , t;
 68         string s;
 69         cin >> a >> s >> b;
 70         t = opToNum.find(s)->second;
 71         if(a == b)
 72             if(t == 1 || t == 5){
 73                 cout << 0;
 74                 return 0;
 75             }
 76             else
 77                 continue;
 78         num[a].push_back(b);
 79         op[a].push_back(t);
 80         num[b].push_back(a);
 81         op[b].push_back(6 - t);
 82         numOp[a]++;
 83         numOp[b]++;
 84     }
 85     ans[0][0] = 1;
 86     for(int i = 1 ; i <= N ; i++){
 87     //DP
 88         for(int j = 2 * i ; j >= 2 ; j--)
 89             if(ans[j - 2][2 * i - j] && ifOK(j , 2 * N - 2 * i + j + 1 , j , j - 1))
 90                 ans[j][2 * i - j] += ans[j - 2][2 * i - j];
 91         for(int j = 2 * i ; j >= 2 ; j--)
 92             if(ans[2 * i - j][j - 2] && ifOK(2 * i - j , 2 * N - j + 1 , 2 * N - j + 1 , 2 * N - j + 2))
 93                 ans[2 * i - j][j] += ans[2 * i - j][j - 2];
 94         for(int j = 1 ; j < 2 * i ; j++)
 95             if(ans[2 * i - j - 1][j - 1] && ifOK(2 * i - j , 2 * N - j + 1 , 2 * i - j , 2 * N - j + 1))
 96                 ans[2 * i - j][j] += ans[2 * i - j - 1][j - 1];
 97     }
 98     int all = 0;
 99     for(int i = 0 ; i <= 2 * N ; i++)
100         all += ans[i][2 * N - i];
101     cout << all / 3;
102     return 0;
103 }